B The NASA Zero Gravity Flight

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Hi,

Please could someone kindly explain how zero weightlessness is achieved in this aircraft. I have tried to use my own understanding and what I have read online.

My ideas are:
To experience weightlessness the plane must be in freefall so the only force acting on it must be gravity. Since planes have wings then simply shutting off the engines will not achieve this since there will still be lift. However, flying in a parabola means that the nose of the plane can tip down over the parabola and effectively remove the lift. On don't understand this picture from wiki though:

Zero_gravity_flight_trajectory_C9-565.jpg



This says that free fall is achieved slightly before and at the peak of the parabola. How does that work? If the plane is perfectly level (like at the top) then there will still be lift surely just like every commercial airliner during the main flight? You would have to effectively tilt the wings downward to get rid of the lift which I thought could only happen after the peak of parabola like I have drawn in green here:

upload_2017-10-1_11-38-19.png


Could someone please explain?

Many thanks!
 
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In the beginning, as in the graph where it says 1.8g you are being accelerated upwards by the airplane. When you reach the 20 second mark the plane stops accelerating by reducing thrust to the point where it exactly matches drag from air resistance, and you continue moving upward at the same velocity. So basically the plane gently falls out from beneath you as if you were a projectile with no air resistance since the plane is isolating you from the wind.
 

jbriggs444

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You would have to effectively tilt the wings downward to get rid of the lift which I thought could only happen after the peak of parabola
You may have noticed that planes are able to maneuver. They do this by pitching up or down relative to the relative airflow. There is no requirement that the plane's nose be aligned with the plane's current path through the air. In order to lose lift, all that is required is for the pilot to push forward on the stick. This points the elevators on the tail downward, pushes the tail upward and pitches the plane downward.

At the top of the parabola, the plane may be pitched slightly downward.

You may also have noticed that the wings on planes are pitched upward. At the top of the parabola, even with the plane pitched somewhat downward, chord of the wings will be closer to horizontal.
 

A.T.

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If the plane is perfectly level (like at the top)
Why does it have to be perfectly level at the top of the parabola? A slight down-pitch is enough to make the total lift of the whole plane zero. The engines then just have to cancel the drag, and you have zero-g.
 

CWatters

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+1

The diagram isn't intended to be super accurate.

Most conventional aircraft have a wing section that produces zero lift at a slightly negative angle of attack. Perhaps only 1 or 2 degrees negative.
 

CWatters

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Just for entertainment I tried to find details for a Boeing 727-200 but could only find polars for the 737. The zero lift angle of attack for the root section was about -1 degree.
 

Andy Resnick

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Hi,

Please could someone kindly explain how zero weightlessness is achieved in this aircraft. I have tried to use my own understanding and what I have read online.

<snip>
Could someone please explain?
I've been lucky enough to have ridden the 'vomit comet' (KC-135 plane) enough to accrue weightlessness time equivalent to a couple of earth orbits.

Everything depends on the pilots (and co-pilots), who are amazingly awesome. Besides 0-g, there are also 'lunars' (0.16 g) and 'martians' (0.38 g). There's a large accelerometer display, and the pilots are able to keep the desired acceleration within a percent or so during the 'reduced-gravity' phase. During a dive, the engine output is reduced, but the plane never 'coasts' because of air resistance.

The diagram you posted isn't exactly right- yes, as the nose comes over the top and diving begins, we enter a short period of weightlessness- however, this period occurs during the dive, not symmetrically as shown in the picture. I know this because we can see out of the windows. Transitions between 'normal' and 'reduced' gravity are rapid- a couple of seconds at most. The climbing period subjects crew members to a 2-g load for a minute or so, that's when most people barf. Overall, the plane executes about 10 parabolas, then climbs during a turn to head the other way- these longer climbs are at lower 'g', maybe 1.5g.
 

A.T.

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Everything depends on the pilots (and co-pilots)
Are these specialized zero-g planes still controlled completely manually, during the maneuver?
 
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I've been lucky enough to have ridden the 'vomit comet' (KC-135 plane) enough to accrue weightlessness time equivalent to a couple of earth orbits.

Everything depends on the pilots (and co-pilots), who are amazingly awesome. Besides 0-g, there are also 'lunars' (0.16 g) and 'martians' (0.38 g). There's a large accelerometer display, and the pilots are able to keep the desired acceleration within a percent or so during the 'reduced-gravity' phase. During a dive, the engine output is reduced, but the plane never 'coasts' because of air resistance.

The diagram you posted isn't exactly right- yes, as the nose comes over the top and diving begins, we enter a short period of weightlessness- however, this period occurs during the dive, not symmetrically as shown in the picture. I know this because we can see out of the windows. Transitions between 'normal' and 'reduced' gravity are rapid- a couple of seconds at most. The climbing period subjects crew members to a 2-g load for a minute or so, that's when most people barf. Overall, the plane executes about 10 parabolas, then climbs during a turn to head the other way- these longer climbs are at lower 'g', maybe 1.5g.
Thanks - very interesting and I'm very jealous :(
 

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